A198949 -id:A198949 - cp's OEIS frontend

A218395 Numbers whose square is the sum of the squares of 11 consecutive integers.

11, 77, 143, 1529, 2849, 30503, 56837, 608531, 1133891, 12140117, 22620983, 242193809, 451285769, 4831736063, 9003094397, 96392527451, 179610602171, 1923018812957, 3583208949023, 38363983731689, 71484568378289, 765356655820823, 1426108158616757

Offset: 0

Paul Weisenhorn, Oct 28 2012

a(n)^2 = Sum_{j=0..10} (x(n)+j)^2 = 11*(x(n)+5)^2 + 110 and b(n) = x(n)+5 give the Pell equation a(n)^2 - 11*b(n)^2 = 110 with the 2 fundamental solutions (11; 1) and (77; 23) and the solution (10; 3) for the unit form. A198949(n+1) = b(n); A106521(n) = x(n) and x(0) = -4.

General: If the sum of the squares of c neighboring numbers is a square with c = 3*k^2-1 and 1 <= k, then a(n)^2 = Sum_{j=0..c-1} (x(n)+j)^2 and b(n) = 2*x(n)+c-1 give the Pell equation a(n)^2 - c*(b(n)/2)^2 = binomial(c+1,3)/2. a(n) = 2*e1*a(n-k) - a(n-2*k); b(n) = 2*e1*b(n-k) - b(n-2*k); a(n) = e1*a(n-k) + c*e2*b(n-k); b(n) = e2*a(n-k) + e1*b(n-k) with the solution (e1; e2) for the unit form.

			For n=6, Sum_{z=17132..17142} z^2 = 3230444569;
a(6) = sqrt(3230444569) = 56837;
b(6) = sqrt((a(6)^2-110)/11) = 17137; x(6) = b(6)-5 = 17132.

V. Pletser, Finding all squared integers expressible as the sum of consecutive squared integers using generalized Pell equation solutions with Chebyshev polynomials, arXiv preprint arXiv:1409.7972 [math.NT], 2014. See Table 1 p. 7.
Index entries for linear recurrences with constant coefficients, signature (0,20,0,-1).

c=2: A001653(n+1) = a(n); A002315(n) = b(n); A001652(n) = x(n).

Cf. A001032 (11 is a term of that sequence), A198947.

s:=0: n:=-1:
for j from -5 to 5 do s:=s+j^2: end do:
for z from -4 to 100000 do
  s:=s-(z-1)^2+(z+10)^2: r:=sqrt(s):
  if (r=floor(r)) then
    n:=n+1: a(n):=r: x(n):=z:
    b(n):=sqrt((s-110)/11):
    print(n,a(n),b(n),x(n)):
  end if:
end do:

LinearRecurrence[{0,20,0,-1},{11,77,143,1529},30] (* Harvey P. Dale, Aug 15 2022 *)

a(n) = 20*a(n-2) - a(n-4); b(n) = 20*b(n-2) - b(n-4);
a(n) = 10*a(n-2) + 33*b(n-2); b(n) = 3*a(n-2) + 10*b(n-2).
a(n) = a(n-1) + 20*a(n-2) - 20*a(n-3) - a(n-4) + a(n-5).
G.f.: 11 * (1-x)*(1+8*x+x^2) / (1 - 20*x^2 + x^4).
With r=sqrt(11); s=10+3*r; t=10-3*r:
a(2*n) = ((11+r)*s^n + (11-r)*t^n)/2.
a(2*n+1) = ((77+23*r) * s^n + (77-23*r)*t^n)/2.
a(n) = 11 * A198947(n+1). - Bill McEachen, Dec 01 2022

A106521 Numbers m such that Sum_{k=0..10} (m+k)^2 is a square.

Original entry on oeis.org

18, 38, 456, 854, 9192, 17132, 183474, 341876, 3660378, 6820478, 73024176, 136067774, 1456823232, 2714535092, 29063440554, 54154634156, 579811987938, 1080378148118, 11567176318296, 21553408328294

Offset: 1

Ralf Stephan, May 30 2005

Equivalently, 11*a(n)^2 + 110*a(n) + 385 is a square.
11*((m+5)^2+10) is a square iff the second factor is divisible by 11 and the quotient is a square, i.e., iff m = 11*k - 4 or m = 11*k - 6 and 11*k^2 +- 2 k + 1 is a square. Thus a(n) == (7,5,5,7,7,5,5,7,...) (mod 11), repeating with period 4 and the values are obtained by solving these Pell-type equations (cf. link to Dario Alpern's quadratic solver). The corresponding recurrence equations (see PARI code) should make it possible to prove the conjectured g.f. - M. F. Hasler, Jan 27 2008
All sequences of this type (i.e., sequences with fixed offset k, and a discernible pattern: k=0...10 for this sequence, k=0..1 for A001652) can be continued using a formula such as x(n) = a*x(n-p) - x(n-2p) + b, where a and b are various constants, and p is the period of the series. Alternatively 'p' can be considered the number of concurrent series. - Daniel Mondot, Aug 05 2016

			Since 18^2 + 19^2 + ... + 28^2 = 5929 = 77^2, 18 is in the sequence. - _Michael B. Porter_, Aug 07 2016

Vincenzo Librandi, Table of n, a(n) for n = 1..500
Dario Alpern, Quadratic two integer variable equation solver
Index entries for linear recurrences with constant coefficients, signature (1,20,-20,-1,1).

Cf. A001032, A094196.

LinearRecurrence[{1,20,-20,-1,1},{18,38,456,854,9192},30] (* Harvey P. Dale, May 07 2011 *)

A106521(n)={local(xy=[ -4-2*(n%2);11],PQRS=[10,3;33,10],KL=[45;165]);until(0>=n-=2,xy=PQRS*xy+KL);xy[1]} \\ M. F. Hasler, Jan 27 2008

G.f.: 2*x*(9+10*x+29*x^2-x^3-2*x^4)/(1-x)/(1-20*x^2+x^4). - Vladeta Jovovic, May 31 2005; adapted to the offset by Bruno Berselli, May 16 2011
a(1)=18, a(2)=38, a(3)=456, a(4)=854, a(5)=9192; thereafter a(n)=a(n-1)+20*a(n-2)- 20*a(n-3)-a(n-4)+a(n-5). - Harvey P. Dale, May 07 2011
a(n) = A198949(n+1)-5. - Bruno Berselli, Feb 12 2012
a(1)=18, a(2)=38, a(3)=456, a(4)=854; thereafter a(n) = 20*a(n-2) - a(n-4) + 90. - Daniel Mondot, Aug 05 2016

Edited and extended by M. F. Hasler, Jan 27 2008

A198947 x values in the solution to 11*x^2 - 10 = y^2.

Original entry on oeis.org

1, 7, 13, 139, 259, 2773, 5167, 55321, 103081, 1103647, 2056453, 22017619, 41025979, 439248733, 818463127, 8762957041, 16328236561, 174819892087, 325746268093, 3487634884699, 6498597125299, 69577877801893, 129646196237887, 1388069921153161, 2586425327632441

Offset: 1

Sture Sjöstedt, Oct 31 2011

When are n and 11*n+1 perfect squares? This problem gives rise to the Diophantine equation 11*x^2 - 10 = y^2.

Positive values of x (or y) satisfying x^2 - 20xy + y^2 + 90 = 0. - Colin Barker, Feb 18 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0, 20, 0, -1).

Cf. A198949, A221762.

m:=26; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(x*(1-x)*(1+8*x+x^2)/(1-20*x^2+x^4))); // Bruno Berselli, Nov 07 2011

LinearRecurrence[{0,20,0,-1},{1,7,13,139},30] (* Vincenzo Librandi, Feb 06 2012 *)

makelist(expand(((11+(-1)^n*sqrt(11))*(10-3*sqrt(11))^floor(n/2)+(11-(-1)^n*sqrt(11))*(10+3*sqrt(11))^floor(n/2))/22), n, 1, 25); /* Bruno Berselli, Nov 07 2011 */

v=vector(25); v[1]=1; v[2]=7; v[3]=13; v[4]=139; for(i=5, #v, v[i]=20*v[i-2]-v[i-4]); v \\ Bruno Berselli, Nov 07 2011

a(n+4) = 20*a(n+2) - a(n) with a(1)=1, a(2)=7, a(3)=13, a(4)=139.
From Bruno Berselli, Nov 06 2011:  (Start)
G.f.: x*(1-x)*(1+8*x+x^2)/(1-20*x^2+x^4).
a(n) = ((11+(-1)^n*t)*(10-3*t)^floor(n/2)+(11-(-1)^n*t)*(10+3*t)^floor(n/2))/22 with t=sqrt(11).  (End).

Terms a(1)-a(7) confirmed, a(8)-a(15) added by John W. Layman, Nov 04 2011

a(16)-a(25) from Bruno Berselli, Nov 06 2011

A221762 Numbers m such that 11*m^2 + 5 is a square.

Original entry on oeis.org

1, 2, 22, 41, 439, 818, 8758, 16319, 174721, 325562, 3485662, 6494921, 69538519, 129572858, 1387284718, 2584962239, 27676155841, 51569671922, 552135832102, 1028808476201, 11015040486199, 20524599852098, 219748673891878, 409463188565759

Offset: 1

Bruno Berselli, Jan 24 2013

Corresponding squares are: 16, 49, 5329, 18496, 2119936, 7360369, 843728209, 2929407376, ... (subsequence of A016778).
The Diophantine equation 11*x^2+k = y^2, for |k|<11, has integer solutions with the following k values:
k = -10, the nonnegative x values are in A198947;
k =  -8,            "                    2*A075839;
k =  -7,            "                    A221763;
k =  -2,            "                    A075839;
k =   1,            "                    A001084;
k =   4,            "                    A075844;
k =   5,            "                    this sequence;
k =   9,            "                    3*A001084.
Also, the Diophantine equation h*x^2+5 = y^2 has infinitely many integer solutions for h = 5, 11, 19, 20, 29, 31, 41, 44, 55, 59, ...
a(n+1)/a(n) tends alternately to (1+sqrt(11))^2/10 and (4+sqrt(11))^2/5.
a(n+2)/a(n) tends to A176395^2/2.

Bruno Berselli, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (0,20,0,-1).

Cf. A001084, A075839, A075844, A198947, A198949, A221763.

Cf. A049629 (numbers m such that 20*m^2 + 5 is a square), A075796 (numbers m such that 5*m^2 + 5 is a square).

m:=24; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+2*x+2*x^2+x^3)/(1-20*x^2+x^4)));

I:=[1,2,22,41]; [n le 4 select I[n] else 20*Self(n-2)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Aug 18 2013

A221762:=proc(q)
local n;
for n from 1 to q do if type(sqrt(11*n^2+5), integer) then print(n);
fi; od; end:
A221762(1000); # Paolo P. Lava, Feb 19 2013

LinearRecurrence[{0, 20, 0, -1}, {1, 2, 22, 41}, 24]
CoefficientList[Series[(1 + 2 x + 2 x^2 + x^3)/(1 - 20 x^2 + x^4), {x, 0, 30}], x] (* Vincenzo Librandi, Aug 18 2013 *)

makelist(expand(((-11*(-1)^n+4*sqrt(11))*(10+3*sqrt(11))^floor(n/2)-(11*(-1)^n+4*sqrt(11))*(10-3*sqrt(11))^floor(n/2))/22), n, 1, 24);

G.f.: x*(1+2*x+2*x^2+x^3)/(1-20*x^2+x^4).
a(n) = -a(1-n) = ((-11*(-1)^n+4*t)*(10+3*t)^floor(n/2)-(11*(-1)^n+4*t)*(10-3*t)^floor(n/2))/22, where t=sqrt(11).
a(n) = 20*a(n-2) - a(n-4) for n>4, a(1)=1, a(2)=2, a(3)=22, a(4)=41.
a(n)*a(n-3)-a(n-1)*a(n-2) = -(3/2)*(9-7*(-1)^n).
a(n+1) + a(n-1) =  A198949(n), with a(0)=-1.
2*a(n-1) - a(n) =  A001084(n/2-1) for even n.

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A218395 Numbers whose square is the sum of the squares of 11 consecutive integers.

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A106521 Numbers m such that Sum_{k=0..10} (m+k)^2 is a square.

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A198947 x values in the solution to 11*x^2 - 10 = y^2.

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A221762 Numbers m such that 11*m^2 + 5 is a square.

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